The *position* as a function of time is 3 cos (wt). The velocity is the derivative of the position with respect to time so -3 w sin(wt) (I used the chain rule and the fact that the derivative of cos is - sin)Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 9. Why is it sin. Time starts from 0 so it should be cos
hmm that was never mentioned in class, and not in lecture notes. He also specified we don't need to know calc, but i get what you are saying. Thanksnrqed said:The *position* as a function of time is 3 cos (wt). The velocity is the derivative of the position with respect to time so -3 w sin(wt) (I used the chain rule and the fact that the derivative of cos is - sin)
Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 13. THis should be it, I have done all the exams
Okay so I found Angular Velocity
W^2 = K/M
K = 72 from the above question
72/8 = W^2
W = 3
T= 2pi/3
t= 2.07.
ohhh ya. Thanksnrqed said:I bet the answer is half of that? Notice that the block will be in contact with the spring for only half the period (from equilibirum position to the farthest to the right and back to equilibirum)
ok but its mass is 2 kg so you could simply use mass over density = 0.02 m^3Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 20. It says Low tide the thing is floats freely on the surface
so what i did was
(1000)(9.8)(X) = (2)(9.8)
X = .002
So .002 is the volume displaced
the Volume of the Entire Cube is .02
? you need to find the buoyant force an dthe force of gravity..002/.02 = .1
which is not the answer
why would there be tension? it is floating freelynrqed said:ok but its mass is 2 kg so you could simply use mass over density = 0.02 m^3
? you need to find the buoyant force an dthe force of gravity.
Then use -Tension +Buoyant force - mg =0
Sorry ! I looked at question 19! SorryAlt+F4 said:why would there be tension? it is floating freely
That's thefraction immersed. They want the fraction above the water lineAlt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 Question 20. It says Low tide the thing is floats freely on the surface
so what i did was
(1000)(9.8)(X) = (2)(9.8)
X = .002
So .002 is the volume displaced
the Volume of the Entire Cube is .02
.002/.02 = .1
which is not the answer
o yaaaa, i think ima take a break for a while. I am losing it after i did 16 exams. Thanksnrqed said:That's thefraction immersed. They want the fraction above the water line
only thing left is just this problemAlt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp01 Question 18
what i did was
pgy = .5p v^2
(1000)(9.8)(.03) = .5(1000)(V^2)
V = .77
Ans: .75
So why am i off by .2
http://online.physics.uiuc.edu/cgi/c...ice/exam3/fa05 Question 24-23. This should be it for a while since ima just do all the exam again
so for this problem i was thinking( i guess similar to the Ballastic one)
pgvdi - (mass of Cylinder + mass of Block) * 9.8 - tension
(1000)(9.8)( .005+ (.2 * .3*.3)) = X+ ( i don't know the mass of block) * 9.8 - 24.5
The .2*.3*.3 is the area of the cube,
Doc Al said:Consider the forces acting on the cube: buoyant force, weight, tension from the string. Since the cube is in equilibrium, those forces must add to zero. Solve for the weight to find the mass.
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa05lightgrav said:This link isn't working ...
If the .2x.3x.3 [m^3] plastic block is pulled down by Tension = 24.5 [N] ,
then you use your first equation to solve for the mass of plastic = 15.5 [kg]
(presuming that speed = 0)
If the small .005 [m^3] cylinder is being pulled upward by Tension,
then the Tension 24.5 [N] enters as a POSitive z-component Force.
Are you DRAWING the Force vectors on your diagrams?
How did you find the KE at 0.01m? Hint: Consider the change in potential energy as the position changes, then use that to find the KE.Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/fa02 14. So i just said K = 2, W = 2 M = 2 and found an imaginary Etotal.
then i found the kinetic energy of .01 and when i divide Etot / 4 i get the kinetic energy that my equation gave. What am i doing wrong
Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp01 Question 18
what i did was
pgy = .5p v^2
(1000)(9.8)(.03) = .5(1000)(V^2)
V = .77
Ans: .75
So why am i off by .2
Sure the marble takes up room in the cup, but which raises the water level more: having the marble imbedded in the ice, or having the marble tossed in the water? (Hint: Which situation displaces more water?)Alt+F4 said:http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam3/sp99 19, absolutely makes no sense. The steel marble will take room in the cup right? thus making the intial level higher unless the water overflows but they never stated that
You are only off by 0.02 m/s. Your method looks OK to me.Alt+F4 said:last question. Any hints?
okay so question 18, i understand that it will be the same as inital level, but then 19 there is an object in there that will take room so it will push the water up. I still don't see why the water would be lower? What happened to the water that was at the intial level?Doc Al said:Sure the marble takes up room in the cup, but which raises the water level more: having the marble imbedded in the ice, or having the marble tossed in the water? (Hint: Which situation displaces more water?)
Why is that? What determines the volume of displaced fluid?Alt+F4 said:okay so question 18, i understand that it will be the same as inital level,
It's not just the fact that "it takes room" that's the key. How about this: Imagine a bathtub of water and a big toy boat floating in it. If I take a bowling ball, which will raise the water level in the bathtub more: (1) dropping the ball in the water, or (2) putting the ball on the boat so that the whole thing floats?but then 19 there is an object in there that will take room so it will push the water up.